Homework: conservation of momentum

Click For Summary

Homework Help Overview

The discussion revolves around a problem in conservation of momentum involving two boats exchanging parcels of mass while traveling in opposite directions. The original poster, Michael, seeks to determine the initial speeds of the boats based on given conditions and masses.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Michael attempts to set up equations based on conservation of momentum and kinetic energy but struggles with the correct application of these principles. Other participants question the assumptions about energy conservation and clarify that this is not an elastic collision.

Discussion Status

Participants are actively engaging with the problem, providing insights into the nature of the collision and discussing the correct equations to use. There is a recognition of the need to express one variable in terms of another to solve the problem, and some participants have confirmed the validity of the equations proposed by Michael.

Contextual Notes

There is an ongoing discussion about the role of kinetic energy in this scenario, with some participants noting that it is not conserved in this inelastic interaction. The original poster expresses confusion about the dependency of the results on the mass of the parcels exchanged.

Michael Kantor
Messages
4
Reaction score
0
Hi,
could you please help me with following:
----------
two boats traveling in opposite direction with speed v1 and v2

when they are passing by they interchange a parcel of same mass M=50kg

as a result second boat stops and first boat continue traveling in same direction with speed u1=8.5m/s

calculate v1 and v2 when you know masses (including M) m1=1000kg and m2=500kg

------------

I wrote
m1*v1+m2*v2=m1*u1

but I am not able to find out second equation. I tried kinetic energy but
I think it is not conserved.

Thank you
Michael
 
Physics news on Phys.org
Welcome to PF,
What do you mean when you say;

Michael Kantor said:
when they are passing by they interchange a parcel of same mass M=50kg

~H
 
Sorry my English is very poor.

one boat is going East and the other West, when they meet
an object of mass 50kg is handed over from boat1 to boat2 and at the same time other object of same mass is handed from boat2 to boat1Thank you
Michael
 
No problem. I would say for this question you have to assume that kinetic energy is conserved. You correctly set up you momentum equation, if you then setup a kinetic energy equation, you have two equations (one for kinetic energy and one for momentum) and two unknowns (v1 and v2), you can then solve both equations simultaneously to obtain v1 and v2.

~H
 
thank you, but ...

Hootenanny said:
No problem. I would say for this question you have to assume that kinetic energy is conserved. You correctly set up you momentum equation, if you then setup a kinetic energy equation, you have two equations (one for kinetic energy and one for momentum) and two unknowns (v1 and v2), you can then solve both equations simultaneously to obtain v1 and v2.

~H

I wrote two equations

v1*m1+v2*m2 = u1*m1

and

(m1*v1^2)/2 + (m2*v2^2)/2 = (m1*u1^2)/2

the result is not M dependent ?
when I solve the equations I get wrong answers.
(correct is v1=9m/s and v2=-1m/s)

Or when and why should M in kinetic energy equation be ?

Thank you very much

Michael
 
Michael Kantor said:
I wrote two equations

v1*m1+v2*m2 = u1*m1

and

(m1*v1^2)/2 + (m2*v2^2)/2 = (m1*u1^2)/2

the result is not M dependent ?
when I solve the equations I get wrong answers.
(correct is v1=9m/s and v2=-1m/s)

Or when and why should M in kinetic energy equation be ?
This is not an elastic collision. Energy is not conserved. Only momentum is conserved. You know that the 50 kg mass traveling at v1 when caught by the second boat causes the second boat to lose all its momentum. Work out the formula for that. That should give you the ratio of the two speeds and will enable you to express v1 in terms of v2.

Then work out the equation for conservaton of momentum for the other boat and substitute your value for v1 in terms of v2. That will give you the answer.

AM
 
I've just worked through the calculations now (I've only just got a spare minute), and I made an incorrect assumtion that the collision was elastic (hence kinetic energy being conserved). If you consider the momentum of the package traveling at v1, you should be able to obtain a ratio of the two velcoities (one will be negative).

~H

Edit: AM got there before me.
 
Last edited:
Andrew Mason said:
This is not an elastic collision. Energy is not conserved.

Just a small niggling point (I'm sure you know, it's just that others might get the wrong idea) - energy (the sum total of all forms) is *always* conserved. Kinetic energy is not conserved (in this case). :smile:
 
Andrew Mason said:
This is not an elastic collision. Energy is not conserved. Only momentum is conserved. You know that the 50 kg mass traveling at v1 when caught by the second boat causes the second boat to lose all its momentum. Work out the formula for that. That should give you the ratio of the two speeds and will enable you to express v1 in terms of v2.

Then work out the equation for conservaton of momentum for the other boat and substitute your value for v1 in terms of v2. That will give you the answer.

AM

The two equations are

M*v1 + (m2-M)*v2 = 0

and

(m1-M)*v1 + M*v2 = m1*u1 ?

I hope so. At least they give correct answer, I use them.:smile:

Thank you
Michael
 
  • #10
Yup, they look correct to me.

~H
 
  • #11
Michael Kantor said:
The two equations are

M*v1 + (m2-M)*v2 = 0

and

(m1-M)*v1 + M*v2 = m1*u1 ?
You could also use:

[tex]m_1v_1 + m_2v_2 = m_1u_1[/tex]

which expresses the conservation of momentum of the whole system.

AM
 
Last edited:

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K